Q115 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
The lines
$\vec{r}=(2 \hat{\jmath}-3 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})$ and $\vec{r}=(2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k})+\mu(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})$ are
- AParallel lines
- Correct answerBIntersecting lines
- CSkew lines
- Dco-incident lines
Solution
The given lines are $\vec{r} = \vec{a}_{1} + \lambda \vec{b}_{1}$ and $\vec{r} = \vec{a}_{2} + \mu \vec{b}_{2}$, where $\vec{a}_{1} = 2\hat{j} - 3\hat{k}$, $\vec{b}_{1} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{a}_{2} = 2\hat{i} + 6\hat{j} + 3\hat{k}$, and $\vec{b}_{2} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
First, check if the lines are parallel. Since $\vec{b}_{1} \times \vec{b}_{2} = (\hat{i} + 2\hat{j} + 3\hat{k}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(8-9) - \hat{j}(4-6) + \hat{k}(3-4) = -\hat{i} + 2\hat{j} - \hat{k} \neq \vec{0}$, the lines are not parallel.
Next, check if the lines intersect by calculating the scalar triple product $(\vec{a}_{2} - \vec{a}_{1}) \cdot (\vec{b}_{1} \times \vec{b}_{2})$.
$\vec{a}_{2} - \vec{a}_{1} = (2\hat{i} + 6\hat{j} + 3\hat{k}) - (2\hat{j} - 3\hat{k}) = 2\hat{i} + 4\hat{j} + 6\hat{k}$.
$(\vec{a}_{2} - \vec{a}_{1}) \cdot (\vec{b}_{1} \times \vec{b}_{2}) = (2\hat{i} + 4\hat{j} + 6\hat{k}) \cdot (-\hat{i} + 2\hat{j} - \hat{k}) = (2)(-1) + (4)(2) + (6)(-1) = -2 + 8 - 6 = 0$.
Since the scalar triple product is zero, the lines are coplanar. As they are not parallel, they must be intersecting lines.
Answer: Intersecting lines
First, check if the lines are parallel. Since $\vec{b}_{1} \times \vec{b}_{2} = (\hat{i} + 2\hat{j} + 3\hat{k}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(8-9) - \hat{j}(4-6) + \hat{k}(3-4) = -\hat{i} + 2\hat{j} - \hat{k} \neq \vec{0}$, the lines are not parallel.
Next, check if the lines intersect by calculating the scalar triple product $(\vec{a}_{2} - \vec{a}_{1}) \cdot (\vec{b}_{1} \times \vec{b}_{2})$.
$\vec{a}_{2} - \vec{a}_{1} = (2\hat{i} + 6\hat{j} + 3\hat{k}) - (2\hat{j} - 3\hat{k}) = 2\hat{i} + 4\hat{j} + 6\hat{k}$.
$(\vec{a}_{2} - \vec{a}_{1}) \cdot (\vec{b}_{1} \times \vec{b}_{2}) = (2\hat{i} + 4\hat{j} + 6\hat{k}) \cdot (-\hat{i} + 2\hat{j} - \hat{k}) = (2)(-1) + (4)(2) + (6)(-1) = -2 + 8 - 6 = 0$.
Since the scalar triple product is zero, the lines are coplanar. As they are not parallel, they must be intersecting lines.
Answer: Intersecting lines
Hence, the correct answer is option: 'B'.